Answer:
0.0793 > 0.01, which means that we have a result in line with the manufacturer's claim.
Step-by-step explanation:
Manufacturer’s claim that the average nicotine content does not exceed 3.5 mg
This means that the null hypothesis is given by:
[tex]H_{0}: \mu = 3.5[/tex]
And the alternate hypothesis is:
[tex]H_{a}: \mu > 3.5[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
3.5 is tested at the null hypothesis
This means that [tex]\mu = 3.5[/tex]
A random sample of 8 cigarettes of a certain brand has an average nicotine content of 4.2 milligrams and a standard deviation of 1.4 milligrams.
This means that [tex]n = 8, X = 4.2, \sigma = 1.4[/tex]
Value of the z-statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{4.2 - 3.5}{\frac{1.4}{\sqrt{8}}}[/tex]
[tex]z = 1.41[/tex]
Pvalue of the test:
We are testing if the mean is higher than 3.5.
The sample mean found is of 4.2, and we have to find the probability of finding a sample mean at least as large as this, which is 1 subtracted by the pvalue of z = 1.41.
z = 1.41 has a pvalue of 0.9207
1 - 0.9207 = 0.0793
0.0793 > 0.01, which means that we have a result in line with the manufacturer's claim.
